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It is known that the group of $K$-rational points of an elliptic curve $E$ is finitely generated if $K$ is a number field of finite degree over $\mathbb{Q}$. Much less is known if $K$ is infinite-dimensional over $\mathbb{Q}$.

The rank of $E$ over $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},...)$ is infinite. It would however suffice for my purposes if I knew that for every prime $p$ there is an infinite family $Q$ of primes, $p\notin Q$, so that:

(1) The rank of $E(\mathbb{Q}(\sqrt{-q}))$ equals that of $E(\mathbb{Q})$ for all $q\in Q$.

(2) $p$ has the same prescribed splitting behavior (i.e. inert, split or ramified) in $\mathbb{Q}(\sqrt{-q})/\mathbb{Q}$ for all $q\in Q$.

Is it known that such a family exists?

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  • $\begingroup$ Not too keen to look for references, but this should be provable. Use some analytic results on the non-vanishing of twists of the $L$-function of $E$ by quadratic characters. There should be a large density of those. (2) then is a congruence condition that can be imposed, too. The result by Kato implies then that the rank of $E$ does not grow in the corresponding quadratic extensions. $\endgroup$
    – Chris Wuthrich
    Commented Jul 13, 2016 at 11:46
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    $\begingroup$ One can obtain what you want if the elements of $Q$ are only required to be squarefree (rather than primes). This follows from results in MR1482805 (say) and the analytic rank 0 case of BSD. If you insist that the elements of $Q$ be primes, then the necessary analytic result can be obtained under GRH by a slight modification of MR1106677 (perhaps it is known unconditionally, but I don't know). $\endgroup$
    – Pasten
    Commented Jul 13, 2016 at 11:54

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Pasten has answered the question: Murty (MS1106677, Corollary to Theorem 2) has shown that the quadratic twist of $E$ by a prime $q$ has rank zero for infinitely many primes $q$, if GRH holds.

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